Question

Sean throws a baseball at an angle of 25 relative to the ground at a speed of 23m/s. What is the maximum height the ball will achieve? How long will it be in the air? How far away does Kelly need to be to catch the ball?

Answer 1

To solve this problem, we can use the equations of motion for projectile motion:

- Vertical displacement: Δy = v₀sinθt - 1/2gt²

- Vertical velocity: v = v₀sinθ - gt

- Time of flight: T = 2v₀sinθ/g

- Horizontal displacement: Δx = v₀cosθt

where v₀ is the initial speed, θ is the angle of projection, g is the acceleration due to gravity (9.81 m/s²), t is the time of flight, and Δy and Δx are the vertical and horizontal displacements, respectively.

First, we can find the maximum height by using the equation for vertical displacement:

Δy = v₀sinθt - 1/2gt²

At the maximum height, the vertical velocity is zero, so we can find the time of flight by setting the vertical velocity equation to zero:

v = v₀sinθ - gt = 0

Solving for t, we get:

t = v₀sinθ/g

Substituting this into the equation for vertical displacement, we get:

Δy = v₀sinθ(v₀sinθ/g) - 1/2g(v₀sinθ/g)²

Simplifying, we get:

Δy = (v₀²sin²θ)/(2g)

Substituting the given values, we get:

Δy = (23²sin²25)/(2*9.81) ≈ 29.4 m

So the maximum height is approximately 29.4 meters.

Next, we can find the time of flight using the equation for time of flight:

T = 2v₀sinθ/g

Substituting the given values, we get:

T = 2*23*sin25/9.81 ≈ 3.5 s

So the time of flight is approximately 3.5 seconds.

Finally, we can find the horizontal displacement using the equation for horizontal displacement:

Δx = v₀cosθt

Substituting the given values, we get:

Δx = 23*cos25*3.5 ≈ 64.3 m

So Kelly needs to be approximately 64.3 meters away to catch the ball.